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A professor says to her class, "You may work these six problems in any order you choose." There are 100 students in the class. Is it possible for each student to work the problems in a different order? Explain.

Short Answer

Expert verified
Yes, each student can work the problems in a different order because there are 720 unique permutations of the six problems, which is more than the 100 students in the class.

Step by step solution

01

Understanding the Problem

The problem involves determining if 100 students can each solve the set of six problems in a unique order. In combinatorics, this relates to the number of permutations of six distinct items.
02

Calculating the Number of Permutations

For n distinct items, the number of different permutations (orders) is given by n factorial, notated as n!. Calculate the number of permutations of the six problems using the factorial function, which for 6! is 6 x 5 x 4 x 3 x 2 x 1.
03

Evaluate the Factorial

Perform the multiplication for 6 factorial: 6! = 6 x 5 x 4 x 3 x 2 x 1 = 720. This is the total number of unique ways to order the six problems.
04

Comparing the Permutations with Number of Students

Since there are 720 unique permutations and only 100 students, it is possible for each student to solve the problems in a different order, as the number of unique orders exceeds the number of students.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Permutations
When we are faced with a set of distinct items and want to figure out in how many different ways we can arrange them, we delve into the concept of permutations. In the context of our exercise, the professor's six problems represent such a set. Understanding permutations is crucial, as they are a fundamental aspect of combinatorics in precalculus.

Permutations allow us to count the number of different ways we can order a set of items without repeating any particular order. This concept is particularly useful in various fields including mathematics, statistics, computer science, and daily decision-making scenarios like arranging books on a shelf or planning a to-do list of tasks.

A key aspect of permutations is that the order matters; 'AB' is different from 'BA.' In mathematical terms, the number of permutations of 'n' distinct items is represented by the notation 'n!' — which is referred to as 'n factorial'. Understanding this notation and the underlying principles is paramount to solving many problems in precalculus.
Factorial Function
The factorial function is represented by an exclamation point (!) and is integral in calculating permutations. The function for any positive integer 'n' (denoted as 'n!') is the product of all positive integers less than or equal to 'n'. For example, '5!' equals 5 × 4 × 3 × 2 × 1. Recognizing how to calculate the factorial is vital in translating the abstract concept of permutations into a solvable numerical problem.

A fascinating characteristic of the factorial function is its rapid growth. As the value of 'n' increases, the value of 'n!' increases exponentially. This is why even for relatively small numbers, like the six problems in our exercise, the factorial function can produce large numbers, explaining the 720 unique permutations available. Another interesting point is that '0!' is defined as 1, which plays a role in certain combinatorial identities and the expansion of binomial coefficients.

Understanding the factorial function lays a strong foundation for not just permutations but also combinations and other concepts in combinatorial mathematics. While dealing with permutations, it's always essential to keep in mind that due to the factorial's nature, the magnitude of potential arrangements can become substantial very quickly.
Problem-Solving Strategies
Tackling combinatorial problems in precalculus can sometimes seem daunting. However, employing effective problem-solving strategies can simplify the process greatly. The first step is to understand the problem: what is being asked and what are the conditions? In our textbook example, our goal is to determine if 100 students can each solve six problems in a unique order.

Once the problem is understood, it's key to identify the right mathematical tools and concepts, in this case, permutations and the factorial function. Breaking down the problem into manageable steps, such as calculating the number of permutations and comparing it to the number of students, helps in progressing towards the solution systematically.

Visualization can be another useful strategy; for instance, picturing the six problems as slots that can be filled helps to comprehend the nature of permutations. Finally, verifying results against the given conditions ensures accuracy. For example, after calculating 720 unique permutations, we check if this number is sufficient for 100 students. If it exceeds the number of students, as it does here, we've confirmed it’s possible for each student to work in a different order. These strategies not just lead to the solution but also aid in building a deeper understanding of the underlying concepts.

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Most popular questions from this chapter

Soccer Team Problem 1: Eleven girls try out for the 11 positions on a soccer team In how many different ways could the 11 positions be filled if there are no restrictions on who plays which position? b. In how many different ways could the positions be filled if Mabel must be the goalkeeper? c. If the positions are selected at random, what is the probability that Mabel will be the goalkeeper? d. What is the probability in part c expressed as a percent?

Admiral Motors manufactures cars with 5 different body styles, 11 different exterior colors, and 6 different interior colors. A dealership wants to display one of each possible variety of car in its showroom. Explain to the manager of the dealership why the plan would be impractical.

Seven people come to an evening bridge party. Only four people can play bridge at any one time, so they decide to play as many games as it takes to use every possible foursome once. How many games would have to be played? Could all of these games be played in one evening?

Baseball Team Problem 2: Nine people try out for the nine positions on a baseball team a. In how many different ways could the positions be filled if there are no restrictions on who plays which position? b. In how many different ways could the positions be filled if Fred must be the pitcher but the other eight people can take any of the remaining eight positions? c. If the positions are selected at random, what is the probability that Fred will be the pitcher? d. What is the probability in part c expressed as a percent?

Evaluate the number of combinations or permutations two ways: a. Using factorials, as in the examples of this section b. Directly, using the built-in features of your grapher $$6 P_{4}$$

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